LeetCode 3100: Water Bottles II Solution

Problem Explanation

Explanation:

The problem is asking to find the maximum number of water bottles that can be drunk given the initial number of full water bottles and the exchange rate.
We can simulate the process by keeping track of the number of full bottles, empty bottles, and the exchange rate.
We continuously exchange empty bottles for full bottles until we do not have enough empty bottles to exchange.
The process stops when the number of empty bottles is fewer than the exchange rate.
The maximum number of water bottles that can be drunk is the sum of the initial full bottles and the number of full bottles obtained from exchanging empty bottles.

Solution Code

class Solution {
    public int numWaterBottles(int numBottles, int numExchange) {
        int total = numBottles;
        int empty = numBottles;
        
        while (empty >= numExchange) {
            int newBottles = empty / numExchange;
            total += newBottles;
            empty = newBottles + empty % numExchange;
        }
        
        return total;
    }
}

Frequently Asked Questions

How to solve LeetCode 3100 (Water Bottles II)?

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The time complexity for LeetCode 3100 (Water Bottles II) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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3100. Water Bottles II